An ideal gas undergoes a process maintaining relation between pressure (P) and volume(V) as P = P₀(1 + (V₀/V)²)⁻¹, where P₀ and V₀ are constants. If two samples A and B (two moles each) with initial volumes V₀ and 3V₀ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, TB - TA is ______. ( R = gas constant) (a) (9P₀V₀)/(8R) (b) (11P₀V₀)/(10R) (c) (7P₀V₀)/(6R) (d) (13P₀V₀)/(11R)

Answer: (b). The key matches the temperature difference at the two initial states, using T = PV/nR with n = 2. A at V₀: P = P₀/2, so TA = ((P₀/2)V₀)/(2R) = (P₀V₀)/(4R). B at 3V₀: P = (P₀)/(1 + 1/9) = (9P₀)/(10), so TB = ((9P₀/10)(3V₀))/(2R) = (27P₀V₀)/(20R). TB - TA = (27 - 5)/(20)(P₀V₀)/(R) = (11P₀V₀)/(10R), so the answer is (b).

An ideal gas undergoes a process maintaining relation between… | JEE Main 2026 (4 Apr, Shift 1)

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Thermodynamic Processes

An ideal gas undergoes a process maintaining relation between pressure $(P)$ and $\text{volume}(V)$ as ${}$

$P = P_0\left(1 + \left(\frac{V_0}{V}\right)^{2}\right)^{-1}$ , where $P_0$ and $V_0$ are constants. If two samples $A$ and $B$ (two moles each) with initial volumes $V_0$ and $3V_0$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_B - T_A$ is ______. ( $R =$ gas constant)

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Answer: (b)

The key matches the temperature difference at the two initial states, using $T = \frac{PV}{nR}$ with $n = 2$ .
A at $V_0$ : $P = \frac{P_0}{2}$ , so $T_A = \frac{(P_0/2)V_0}{2R} = \frac{P_0V_0}{4R}$ .
B at $3V_0$ : $P = \frac{P_0}{1 + 1/9} = \frac{9P_0}{10}$ , so $T_B = \frac{(9P_0/10)(3V_0)}{2R} = \frac{27P_0V_0}{20R}$ .
$T_B - T_A = \frac{27 - 5}{20}\frac{P_0V_0}{R} = \frac{11P_0V_0}{10R}$ , so the answer is (b).

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